Topic 16
Interval Estimation
Our strategy to estimation thus far has been to use a method to find an estimator, e.g., method of moments, or maximum likelihood, and evaluate the quality of the estimator by evaluating the bias and the variance of the estimator. Often, we know more about the distribution of the estimator and this allows us to take a more comprehensive statement about the estimation procedure. Interval estimation is an alternative to the variety of techniques we have examined. Given data x, we replace the point estimate ✓ˆ(x) for the parameter ✓ by a statistic that is subset Cˆ(x) of the parameter space. We will consider both the classical and Bayesian approaches to choosing Cˆ(x) . As we shall learn, the two approaches have very different interpretations.
16.1 Classical Statistics
In this case, the random set Cˆ(X) is chosen to have a prescribed high probability, , of containing the true parameter value ✓. In symbols, P ✓ Cˆ(X) = . ✓{ 2 } In this case, the set Cˆ(x) is called a -level con-
fidence set. In the case of a one dimensional param- 0.4 eter set, the typical choice of confidence set is a con-
fidence interval 0.35
Cˆ(x)=(✓ˆ`(x), ✓ˆu(x)). 0.3 Often this interval takes the form 0.25 Cˆ(x)=(✓ˆ(x) m(x), ✓ˆ(x)+m(x)) = ✓ˆ(x) m(x) ± where the two statistics, 0.2 ✓ˆ(x) is a point estimate, and • 0.15 m(x) is the margin of error. • 0.1
16.1.1 Means 0.05 area α Example 16.1 (1-sample z interval). If X1.X2....Xn 0 µ z are normal random variables with unknown mean −3 −2 −1 0 1 2 α 3 2 but known variance 0. Then, . ↵ is the area under the standard X¯ µ Figure 16.1: Upper tail critical values Z = normal density and to the right of the vertical line at critical value z↵ 0/pn
239 Introduction to the Science of Statistics Interval Estimation
is a standard normal random variable. For any ↵ between 0 and 1, let z↵ satisfy P Z>z = ↵ or equivalently P Z z =1 ↵. { ↵} { ↵} The value is known as the upper tail probability with critical value z↵. We can compute this in R using, for example > qnorm(0.975) [1] 1.959964 for ↵ =0.025. If =1 2↵, then ↵ =(1 )/2. In this case, we have that P z